Answer:
P(exactly two buy yearbooks) = 0.31104
P(at least two buy yearbooks) = 0.76672
Step-by-step explanation:
We are given that at high school 40% of the students buy yearbooks.
You select 6 students at random.
The above situation can be represented through binomial distribution;
[tex]P(X = r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r};x=0,1,2,3,.......[/tex]
where, n = number trials (samples) taken = 6 students
r = number of success
p = probability of success which in our question is probability that
students buy yearbooks, i.e; p = 0.40
Let X = Number of students who buy yearbooks
So, X ~ Binom(n = 6, p = 0.40)
(a) Now, Probability that exactly two buy yearbooks is given by = P(X = 2)
P(X = 2) = [tex]\binom{6}{2} \times 0.40^{2} \times (1-0.40)^{6-2}[/tex]
= [tex]15\times 0.40^{2} \times 0.60^{4}[/tex]
= 0.31104
(b) Probability that at least two buy yearbooks is given by = P(X [tex]\geq[/tex] 2)
P(X [tex]\geq[/tex] 2) = 1 - P(X = 0) - P(X = 1)
= [tex]1- \binom{6}{0} \times 0.40^{0} \times (1-0.40)^{6-0}-\binom{6}{1} \times 0.40^{1} \times (1-0.40)^{6-1}[/tex]
= [tex]1- (1 \times 1 \times 0.60^{6})-(6 \times 0.40^{1} \times 0.60^{5})[/tex]
= 1 - 0.04666 - 0.18662
= 0.76672
